Let us assume that the variables $u_{1}$ and $u_{2}$ vary in such a way that their sum $u_{1} + u_{2}$ is a constant. And we set: \begin{align} s_{1} = \text{sn}(u_{1}, k),\, c_{1} = \text{cn}&(u_{1}, k),\, d_{1} = \text{dn}(u_{1}, k)\notag\\ s_{2} = \text{sn}(u_{2}, k),\, c_{2} = \text{cn}&(u_{2}, k),\, d_{2} = \text{dn}(u_{2}, k)\notag\\ u_{1} + u_{2} &= a\notag \end{align} so that $$\frac{du_{2}}{du_{1}} = -1$$ Then we have \begin{align} \frac{d}{du_{1}}(s_{1} + s_{2}) &= c_{1}d_{1} - c_{2}d_{2}\notag\\ \frac{d}{du_{1}}(s_{1}c_{2} + s_{2}c_{1}) &= s_{1}s_{2}d_{2} + c_{1}d_{1}c_{2} - s_{2}s_{1}d_{1} - c_{2}d_{2}c_{1}\notag\\ &= (d_{1} - d_{2})(c_{1}c_{2} - s_{1}s_{2})\notag\\ \frac{d}{du_{1}}(d_{1} + d_{2}) &= -k^{2}(s_{1}c_{1} - s_{2}c_{2})\notag\\ &= -k^{2}s_{1}c_{1}(c_{2}^{2} + s_{2}^{2}) + k^{2}s_{2}c_{2}(c_{1}^{2} + s_{1}^{2})\notag\\ &= -k^{2}(c_{1}c_{2} - s_{1}s_{2})(s_{1}c_{2} - s_{2}c_{1})\notag \end{align} Since we have \begin{align} -k^{2}(s_{1}^{2}c_{2}^{2} - s_{2}^{2}c_{1}^{2}) &= -k^{2}(s_{1}^{2}(1 - s_{2}^{2}) - s_{2}^{2}(1 - s_{1}^{2}))\notag\\ &= -k^{2}(s_{1}^{2} - s_{2}^{2})\notag\\ &= d_{1}^{2} - d_{2}^{2}\notag \end{align} It therefore follows that $$(d_{1} + d_{2})\frac{d}{du_{1}}(s_{1}c_{2} + s_{2}c_{1}) = (s_{1}c_{2} + s_{2}c_{1})\frac{d}{du_{1}}(d_{1} + d_{2})$$ From the above it is clear that the ratio $(s_{1}c_{2} + s_{2}c_{1})/(d_{1} + d_{2})$ is a constant independent of $u_{1}$ as long as the sum $u_{1} + u_{2} = a$ is a constant. Therefore value of this ratio is obtained by putting $u_{1} = 0$ and $u_{2}= a$ and we have $$\frac{s_{1}c_{2} + s_{2}c_{1}}{d_{1} + d_{2}} = \frac{\text{sn}(a, k)}{1 + \text{dn}(a, k)} = \frac{\text{sn}(u_{1} + u_{2}, k)}{1 + \text{dn}(u_{1} + u_{2}, k)}$$ Again we can see that \begin{align} \frac{s_{1}c_{2} - s_{2}c_{1}}{d_{1} - d_{2}} &= -\frac{1}{k^{2}}\cdot \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}} = \text{ a constant independent of }u_{1}\notag\\ &= \frac{\text{sn}(a, k)}{\text{dn}(a, k) - 1}\text{ by putting } u_{1} = a, u_{2} = 0\notag\\ &= \frac{\text{sn}(u_{1} + u_{2}, k)}{\text{dn}(u_{1} + u_{2}, k) - 1}\notag \end{align} We therefore have the following results: \begin{align} \frac{\text{dn}(u_{1} + u_{2}, k) + 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}}\tag{A}\\ \frac{\text{dn}(u_{1} + u_{2}, k) - 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{d_{1} - d_{2}}{s_{1}c_{2} - s_{2}c_{1}}\tag{B} \end{align} Subtracting the above equations we get $$\frac{2}{\text{sn}(u_{1} + u_{2}, k)} = \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}} - \frac{d_{1} - d_{2}}{s_{1}c_{2} - s_{2}c_{1}} = \frac{2(s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1})}{s_{1}^{2}c_{2}^{2} - s_{2}^{2}c_{1}^{2}}$$ and therefore $$\text{sn}(u_{1} + u_{2}, k) = \frac{s_{1}^{2} - s_{2}^{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}$$ Adding the equations $(A)$ and $(B)$ we get $$2\text{ds}(u_{1} + u_{2}, k) = \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}} + \frac{d_{1} - d_{2}}{s_{1}c_{2} - s_{2}c_{1}} = \frac{2(s_{1}c_{2}d_{1} - s_{2}c_{1}d_{2})}{s_{1}^{2} - s_{2}^{2}}$$ and therefore $$\text{dn}(u_{1} + u_{2}, k) = \frac{s_{1}c_{2}d_{1} - s_{2}c_{1}d_{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}$$ In similar fashion one can obtain the following formulas: \begin{align} \frac{\text{cn}(u_{1} + u_{2}, k) + 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{c_{1} + c_{2}}{s_{1}d_{2} + s_{2}d_{1}}\notag\\ \frac{\text{cn}(u_{1} + u_{2}, k) - 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{c_{1} - c_{2}}{s_{1}d_{2} - s_{2}d_{1}}\notag \end{align} Adding the above equations we get $$2\text{cs}(u_{1} + u_{2}, k) = \frac{2(s_{1}c_{1}d_{2} - s_{2}c_{2}d_{1})}{s_{1}^{2} - s_{2}^{2}}$$ and finally we have $$\text{cn}(u_{1} + u_{2}, k) = \frac{s_{1}c_{1}d_{2} - s_{2}c_{2}d_{1}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}$$ To summarize we have the following addition formulas: \begin{align} \text{sn}(u_{1} + u_{2}, k) &= \frac{s_{1}^{2} - s_{2}^{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}\tag{1}\\ \text{cn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{1}d_{2} - s_{2}c_{2}d_{1}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}\tag{2}\\ \text{dn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{2}d_{1} - s_{2}c_{1}d_{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}\tag{3} \end{align} The above forms are hard to remember and don't look similar to addition formulas for circular functions. However they can be transformed by algebraic manipulations into a form which is similar to those of circular functions. This we will do in the next post. For the time being we can use these results to obtain expressions for $\text{sn}(u + K, k),\, \text{cn}(u + K, k),\, \text{dn}(u + K, k)$.
Putting $u_{1} = u$ and $u_{2} = K$ we get \begin{align} \text{sn}(u + K, k) &= \frac{\text{cn}(u, k)}{\text{dn}(u, k)} = \text{cd}(u, k)\tag{4}\\ \text{cn}(u + K, k) &= \frac{-k^{\prime}\,\text{sn}(u, k)}{\text{dn}(u, k)} = -k^{\prime}\,\text{sd}(u, k)\tag{5}\\ \text{dn}(u + K, k) &= \frac{k^{\prime}}{\text{dn}(u, k)} = k^{\prime}\,\text{nd}(u, k)\tag{6} \end{align} We can now put $u_{1} + K$ in place of $u_{1}$ in the addition formulas to get \begin{align} \text{sn}(u_{1} + K + u_{2}, k) &= \dfrac{\dfrac{c_{1}^{2}}{d_{1}^{2}} - s_{2}^{2}}{\dfrac{c_{1}c_{2}d_{2}}{d_{1}} + \dfrac{k'^{2}s_{1}s_{2}}{d_{1}^{2}}}\notag\\ \Rightarrow \frac{\text{cn}(u_{1} + u_{2}, k)}{\text{dn}(u_{1} + u_{2}, k)} &= \frac{c_{1}^{2} - d_{1}^{2}s_{2}^{2}}{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}\notag \end{align} and \begin{align} \text{dn}(u_{1} + K + u_{2}, k) &= \dfrac{\dfrac{k'c_{1}c_{2}}{d_{1}^{2}} + k'\dfrac{s_{1}s_{2}d_{2}}{d_{1}}}{\dfrac{c_{1}c_{2}d_{2}}{d_{1}} + \dfrac{k'^{2}s_{1}s_{2}}{d_{1}^{2}}}\notag\\ \Rightarrow \frac{k'}{\text{dn}(u_{1} + u_{2}, k)} &= \frac{k'c_{1}c_{2} + k's_{1}s_{2}d_{1}d_{2}}{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}\notag\\ \Rightarrow \text{dn}(u_{1} + u_{2}, k) &= \frac{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\notag\\ \Rightarrow \text{cn}(u_{1} + u_{2}, k) &= \frac{c_{1}^{2} - d_{1}^{2}s_{2}^{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}} = \frac{1 - s_{1}^{2} - s_{2}^{2} + k^{2}s_{1}^{2}s_{2}^{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\notag \end{align} Again \begin{align} \text{cn}(u_{1} + K + u_{2}, k) &= \dfrac{-\dfrac{k's_{1}c_{1}d_{2}}{d_{1}^{2}} - \dfrac{k's_{2}c_{2}}{d_{1}}}{\dfrac{c_{1}c_{2}d_{2}}{d_{1}} + \dfrac{k'^{2}s_{1}s_{2}}{d_{1}^{2}}}\notag\\ \Rightarrow \frac{\text{sn}(u_{1} + u_{2}, k)}{\text{dn}(u_{1} + u_{2}, k)} &= \frac{s_{1}c_{1}d_{2} + s_{2}c_{2}d_{1}}{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}\notag \end{align} Thus we can now write the addition formulas in a different form: \begin{align} \text{sn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{1}d_{2} + s_{2}c_{2}d_{1}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\tag{7}\\ \text{cn}(u_{1} + u_{2}, k) &= \frac{1 - s_{1}^{2} - s_{2}^{2} + k^{2}s_{1}^{2}s_{2}^{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\tag{8}\\ \text{dn}(u_{1} + u_{2}, k) &= \frac{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\tag{9} \end{align} We have so far developed the addition formulas in two different forms and we note that in each of the forms the formulas have a common denominator. Using them we have also obtained the expressions for $\text{sn}(u + K, k)$ etc.